Question -
Answer -
Given:
cos A = -12/13 and cot B = 24/7
We know that, A lies in second quadrant, B in the third quadrant.
In the second quadrant sine function is positive.
In the third quadrant, both sine and cosine functions are negative.
By using the formulas,
sin A = √(1 – cos2 A), sin B = – 1/√(1 + cot2 B) and cos B = -√(1 – sin2 B),
So let us find the value of sin A and sin B
sin A = √(1 – cos2 A)
= √(1 – (-12/13)2)
= √(1 – 144/169)
= √((169-144)/169)
= √(25/169)
= 5/13
sin B = – 1/√(1 + cot2 B)
= – 1/√(1 + (24/7)2)
= – 1/√(1 + 576/49)
= -1/√((49+576)/49)
= -1/√(625/49)
= -1/(25/7)
= -7/25
cos B = -√(1 – sin2 B)
= -√(1-(-7/25)2)
= -√(1-(49/625))
= -√((625-49)/625)
= -√(576/625)
= -24/25
So, now let us find
(i) sin (A + B)
We know that sin (A + B) = sin A cos B + cos A sin B
So,
sin (A + B) = sin A cos B + cos A sin B
= 5/13 × (-24/25) + (-12/13) × (-7/25)
= -120/325 + 84/325
= -36/325
(ii) cos (A + B)
We know that cos (A + B) = cos A cos B – sin A sin B
So,
cos (A + B) = cos A cos B – sin A sin B
= -12/13 × (-24/25) – (5/13) × (-7/25)
= 288/325 + 35/325
= 323/325
(iii) tan (A + B)
We know that tan (A + B) = sin (A+B) / cos (A+B)
= (-36/325) / (323/325)
= -36/323